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978-0134610672 Chapter 1 Part 1
1 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by […]
978-0134610672 Chapter 1 Part 2
1 4 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or […]
978-0134610672 Chapter 10 Part 1
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978-0134610672 Chapter 10 Part 2
500 SOLUTION 1 FEM 2 BC = 1 FEM 2 CD = –21621152 9 =–20 k # f t 1FEM2CB =1FEM2DC =20 k #ft M N=3E aI Lb1 uN–c 2 + 1 FEM 2N M BA = 3EI 12 1 […]
978-0134610672 Chapter 11 Part 1
518 SOLUTION ( DF)AB =0 (DF)BA = 0.75I BC >24 0.75IBC>24 +IBC>16 =0.333 3 (DF)BC =0.6667 (DF)CB =0 ( FEM)AB = –0.8(24)2 12 =–38.4 k #f t (FEM)BA =38.4 k #ft ( FEM)BC =– 30(16) 8 =–60.0 k # f […]
978-0134610672 Chapter 11 Part 2
538 SOLUTION FEMAC =0 FEMBD =0 F EMCD =–wL 2 12 =– 6(82) 12 =–32 kN #m=–FEMD C F EMDE =–wL 2 8 =– 6(72) 8 =–36.75 kN # m Joint A B C D E Member AC BD CA […]
978-0134610672 Chapter 12 Part 1
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978-0134610672 Chapter 12 Part 2
578 SOLUTION Method of Sections. It is required that FCE =FDF =F1. Referring to Fig. a, + SΣFx=0; 8–2F1 a3 5b =0 F1=6.667 k N Therefore, FCE =6.67 kN (C) FDF =6.67 kN (T) Ans. a +ΣME=0; FCD(1.5) –6.667 a4 […]
978-0134610672 Chapter 12 Part 3
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 12 Part 4
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 13
632 Solution a A= 6 20 =0.3 aB= 4 20 =0.2 r A=rB= 4–2 2 = 1 From Table 13–1, for span AB, CAB =0.622 CBA =0.748 KAB =10.06 KBA =8.37 K BA = K BA EI C L = […]
978-0134610672 Chapter 14 Part 1
649 SOLUTION Member 1: l x= 1 – 0 12 =0.7071 ly= 1 – 2 12 =–0.707 1 k 1=AE ≥ 1 2 3 4 0.3536 –0.3536 –0.3536 0.3536 –0.3536 0.3536 0.3536 –0.3536 –0.3536 0.3536 0.3536 –0.3536 0.3536 –0.3536 –0.3536 […]
978-0134610672 Chapter 14 Part 2
660 SOLUTION The origin of the global coordinate system is set at joint 1 . For member 2 , L=5 m. Referring to Fig. a, u x″=180°–45°– sin –1 a4 5b =81.87 ° uy″=171.87°. Thus, lx″= cos ux″= cos 81.87°=0.14142 […]
978-0134610672 Chapter 15
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978-0134610672 Chapter 16 Part 1
687 SOLUTION Member 1. l x= 10 –0 10 =1 ly= 0 AE L = 20(29)(103) 10(12) =4833.33 k > in. 12EI L 3= 12(29)(103)(650) (10) 3 (12) 3=130.90 k > in . 6EI L 2= 6(29)(103)(650) (10) 2 (12) […]
978-0134610672 Chapter 16 Part 2
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 2 Part 1
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 2 Part 2
4 6 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or […]
978-0134610672 Chapter 2 Part 3
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 3 Part 1
7 2 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or […]
978-0134610672 Chapter 3 Part 2
9 2 Ans. FCE =FEG =FEF =0 FDE =2.00 k 1C2 FFG =3.00 k 1C2 FAG =2.00 k 1C2 FCD =4.375 k 1C2 FAD =2.625 k 1T2 FCF =4.375 k 1C2 FBF =0.625 k 1T2 FAB =0.625 k 1T2 Equations […]
978-0134610672 Chapter 3 Part 3
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 4 Part 1
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 4 Part 2
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 4 Part 3
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 4 Part 4
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 5 Part 1
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 5 Part 2
208 5–21. The power transmission cable weighs 10 lb>ft . If the resultant horizontal force on tower BD is required to be zero, determine the sag h of cable BC. AB hC D 300 ft 10 ft 200 ft SOLUTION […]
978-0134610672 Chapter 6 Part 1
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 6 Part 2
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 6 Part 3
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 6 Part 4
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 6 Part 5
300 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by […]
978-0134610672 Chapter 7 Part 1
313 7–1. Determine the equation of the elastic curve using the coordinate x, and specify the slope at point A and the deflection at point C. EI is constant. SOLUTION Support Reactions and Elastic Curve: As shown on FBD(a). Moment […]
978-0134610672 Chapter 7 Part 2
333 Ans. a=0.153 L © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any […]
978-0134610672 Chapter 7 Part 3
346 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by […]
978-0134610672 Chapter 8 Part 1
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 8 Part 2
379 8–21. Determine the slope and displacement at point C. Use the principle of virtual work. EI is constant. A 2 m1 m B C 3 m 12 kN > m SOLUTION Referring to the virtual moment functions shown in […]
978-0134610672 Chapter 8 Part 3
399 *8–40. Use the method of virtual work and determine the slope at point A of the beam made from steel. E=29(103) ksi, I=245 in4. 5 ft 8 k 5 ft AB 8 k 5 ft 5 ft C SOLUTION […]
978-0134610672 Chapter 8 Part 4
© 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any […]
978-0134610672 Chapter 9 Part 1
421 9–1. Determine the reactions at the supports, then draw the shear and moment diagrams. Assume the support at A is fixed and B is a roller. EI is constant. B A 12 ft 500 lb / ft SOLUTION Support […]
978-0134610672 Chapter 9 Part 2
441 Ans. NA=15.6 kN Cx=0 Cy=38.4 kN MC=14.4 kN #m © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this […]
978-0134610672 Chapter 9 Part 3
461 © 2018 Pearson Education, Inc., 330 Hudson Street, NY, NY 10013. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by […]
978-0134610672 Chapter 9 Part 4
472 SOLUTION The primary real beam and qualitative influence line are shown in Fig. a and its conjugate beam is shown in Fig. b. Referring to Fig. c, aAA = EI , fAA =M′ A=0, fBA =M′ B=0, fCA =M′ […]